Perplexity AI Investor Presentation Deck

September 27: ChatGPT like Slackbot Aravind Srinivas 12 months ago @Perplexity.al I have a table named "users". users[id.name.email). My table structure is as follows: ID is primary key auto increment I would like to return the email, count and the last_id of user that has the same email address TABLE : Johnny Ho 12:04 AM @Perplexity.al tell me about the most popular vector dbs P4 replies Last reply 12 months ago | Johnny Ho 12:05 AM @Perplexity.al tell me about fast knn search databases 2 replies Last reply 12 months ago Johnny Ho 3:54 PM @Perplexity.al what is SOQL? replies Last reply 12 months ago September 27th, 2022 Aravind Srinivas 4:13PM @Perplexity.al how to prepend "image-" to all filenames of png format 2 replies Last reply 12 months ago Aravind Srinivas 4:14 PM @Perplexity.al bash command to prepend "image-" to all files of png format 2 replies Last reply 12 months ago Aravind Srinivas 4:14 PM Perplexity.al how to start a server in ruby 2 replies Last reply 12 months ago Aravind Srinivas 4:15 PM @Perplexity.al bash command for starting a static server with ruby 2 replies Last reply 12 months ago Aravind Srinivas 4:15 PM Perplexity.al bash command for starting a static server with php 2 replies Last reply 12 months ago In other words, my query should return last_id: 3-count: 2 - email: Jonesones.com last_id: 7 - count: 3-email: [email protected] I know a piece of the query that returns the count, and the email that is repeated, but I can't get the last id. SELECT email.countlemail) FROM users GROUP BY email HAVING COUNT(email) > 1 ORDER BY countlemail) DESC 2 replies P Perplexity.al A 12 months ago Data source chat Show preview Open in Streamlit Perplexity.al APP 12 months ago The best way to get the last ID of a user with the same email address would be to use a subquery. Something like this should work SELECT email, count(email), (SELECT id FROM users WHERE emailu.email ORDER BY id DESC LIMIT 1) as last_id FROM users u GROUP BY email HAVING COUNT(email) > 1 ORDER BY count(email) DESC See less

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